Calc-Ok Question 1
Most-appropriate topic codes (CED):
• TOPIC 2.3: Estimating Derivatives of a Function at a Point — part (b)
• TOPIC 8.2: Connecting Position, Velocity, and Acceleration of Functions Using Integrals — part (c)
• TOPIC 4.6: Approximating Values Using Local Linearity and Linearization — part (d)
▶️ Answer/Explanation
Average rate of change on \([0,30]\):
\(\displaystyle \frac{A(30)-A(0)}{30-0} =\frac{6.687(0.931)^{30}-6.687}{30} \approx \frac{0.7829-6.6870}{30}\)
\(\displaystyle \approx -0.1968\ \text{pounds/day}.\)
(b)
\(A'(t)=6.687(0.931)^{t}\ln(0.931)=A(t)\ln(0.931).\)
\(\displaystyle A'(15)=A(15)\ln(0.931)\approx 2.2881\times(-0.071496)\approx -0.1636\ \text{pounds/day}.\)
Interpretation: At \(t=15\) days, the amount of grass clippings is decreasing at about \(0.164\) pound per day.
(c)
Average amount on \([0,30]\): \(\displaystyle \bar A=\frac{1}{30}\int_{0}^{30}A(t)\,dt.\)
\(\displaystyle \int A(t)\,dt=\int 6.687(0.931)^{t}\,dt =\frac{6.687}{\ln(0.931)}(0.931)^{t}+C.\)
Thus \(\displaystyle \bar A=\frac{1}{30}\cdot \frac{6.687}{\ln(0.931)}\Big[(0.931)^{30}-1\Big].\)
Set \(A(t)=\bar A\): \(6.687(0.931)^{t}=\bar A\Rightarrow (0.931)^{t}=\dfrac{(0.931)^{30}-1}{30\,\ln(0.931)}.\)
Solve: \(\displaystyle t=\frac{\ln\!\left(\dfrac{(0.931)^{30}-1}{30\,\ln(0.931)}\right)}{\ln(0.931)}\approx \boxed{12.415\ \text{days}}.\)
(d)
Linearization at \(t=30\): \(\displaystyle L(t)=A(30)+A'(30)\,(t-30).\)
\(A(30)\approx 0.7829,\quad A'(30)=A(30)\ln(0.931)\approx 0.7829\times(-0.071496)\approx -0.05598.\)
Set \(L(t)=0.5\): \(0.7829+(-0.05598)(t-30)=0.5.\)
\(\displaystyle t-30=\frac{0.5-0.7829}{-0.05598}\approx 5.054\ \Rightarrow\ \boxed{t\approx 35.054\ \text{days}}.\)
Calc-Ok Question 2
Most-appropriate topic codes (CED):
• TOPIC 9.5: Derivatives in Polar/Cartesian Forms — part (b)
• TOPIC 9.6: Rates and Motion with Polar/Parametric Functions — parts (c), (d)
▶️ Answer/Explanation
Inside both curves: \(0\le r\le \min\{3,\ 3-2\sin(2\theta)\}\).
On \(0\le\theta\le\dfrac{\pi}{2}\): \(3-2\sin(2\theta)\le3\) ⇒ use \(r=3-2\sin(2\theta)\).
On \(\dfrac{\pi}{2}\le\theta\le\pi\): \(3\) is smaller.
\[ \text{Area} =\tfrac12\!\int_{0}^{\pi/2}\!(3-2\sin(2\theta))^{2}\,d\theta +\tfrac12\!\int_{\pi/2}^{\pi}\!3^{2}\,d\theta. \] Expand: \((3-2\sin(2\theta))^{2}=11-12\sin(2\theta)-2\cos(4\theta)\).
First piece: \(\displaystyle \int_{0}^{\pi/2}(11-12\sin(2\theta)-2\cos(4\theta))\,d\theta=\frac{11\pi}{2}-12\).
Half of that: \(\displaystyle \frac{11\pi}{4}-6\).
Second piece: \(\displaystyle \tfrac12\!\int_{\pi/2}^{\pi}9\,d\theta=\frac{9\pi}{4}\).
Total: \(\boxed{5\pi-6}\approx 9.708.\)
(b) \(\dfrac{dx}{d\theta}\) at \(\theta=\dfrac{\pi}{6}\)
\(x=r\cos\theta\Rightarrow \dfrac{dx}{d\theta}=\dfrac{dr}{d\theta}\cos\theta – r\sin\theta.\)
\(r=3-2\sin(2\theta)\Rightarrow \dfrac{dr}{d\theta}=-4\cos(2\theta).\)
At \(\theta=\dfrac{\pi}{6}\): \(r=3-\sqrt{3},\ \cos\theta=\tfrac{\sqrt{3}}{2},\ \sin\theta=\tfrac12,\ \cos(2\theta)=\tfrac12\).
\(\displaystyle \frac{dx}{d\theta} =(-4\times\tfrac12)\times\tfrac{\sqrt{3}}{2}-(3-\sqrt{3})\times\tfrac12 =-\frac{3+\sqrt{3}}{2}\approx -2.366.\)
(c) Rate of change of distance at \(\theta=\dfrac{\pi}{3}\)
For \(0<\theta<\dfrac{\pi}{2}\), distance between curves along a ray is \(d(\theta)=3-(3-2\sin(2\theta))=2\sin(2\theta).\)
\(\displaystyle \frac{dd}{d\theta}=4\cos(2\theta).\) At \(\theta=\dfrac{\pi}{3}\): \(\cos(2\theta)=\cos\!\left(\dfrac{2\pi}{3}\right)=-\tfrac12\).
\(\boxed{\dfrac{dd}{d\theta}=-2}\) (units per radian).
(d) \(\dfrac{dr}{dt}\) at \(\theta=\dfrac{\pi}{6}\) when \(\dfrac{d\theta}{dt}=3\)
\(\displaystyle \frac{dr}{dt}=\frac{dr}{d\theta}\times\frac{d\theta}{dt} =\big(-4\cos(2\theta)\big)\times 3=-12\cos(2\theta).\)
At \(\theta=\dfrac{\pi}{6}\): \(\cos(2\theta)=\tfrac12\).
\(\boxed{\dfrac{dr}{dt}=-6}\) (radius decreasing at 6 units per time).
No-Calc Question 3
The graph of \(f\) consists of three line segments and is shown above.
Let \(g\) be the function defined by \(g(x)=\displaystyle\int_{-3}^{x} f(t)\,dt\).
Give a reason for your answer.
Find the slope of the line tangent to the graph of \(p\) at the point where \(x=-1\).
Most-appropriate topic codes (CED):
• TOPIC 5.3: Determining Intervals on Which a Function Is Increasing or Decreasing — part (b)
• TOPIC 2.9: The Quotient Rule — part (c)
• TOPIC 2.8: Product/Chain Rule Applications — part (d)
▶️ Answer/Explanation
(a)
\(g(3)=\displaystyle\int_{-3}^{3} f(t)\,dt\).
From the graph, split into pieces:
\(-3\to 0\): triangle area \(=\tfrac12(3)(4)=6\).
\(0\to 2\): triangle area \(=\tfrac12(2)(4)=4\).
\(2\to 3\): triangle (below axis) \(=-\tfrac12(1)(2)=-1\).
So \(g(3)=6+4-1=\boxed{9}.\)
(b)
\(g'(x)=f(x)\), \(g”(x)=f'(x)\).
Increasing & concave down \(\Longleftrightarrow\) \(f(x)>0\) and \(f'(x)<0\).
From the graph, this occurs on \(\boxed{(-5,-3)\ \text{ and }\ (0,2)}\).
(c)
\(h(x)=\dfrac{g(x)}{5x}\). Use the Quotient Rule:
\(h'(x)=\dfrac{5x\,g'(x)-5\,g(x)}{(5x)^{2}}\).
At \(x=3\): \(g'(3)=f(3)=-2,\ g(3)=9\).
\(h'(3)=\dfrac{(5\cdot 3)(-2)-5\cdot 9}{25\cdot 3^{2}} =\dfrac{-30-45}{225} =\boxed{-\tfrac13}.\)
(d)
\(p(x)=f(x^{2}-x)\). Chain Rule:
\(p'(x)=f'(x^{2}-x)\,(2x-1).\)
At \(x=-1\): inner \(=2\); slope of the segment at \(x=2\) is \(-2\Rightarrow f'(2)=-2\).
\(p'(-1)=(-2)\,(-3)=\boxed{6}.\)
No-Calc Question 4
| t (minutes) | 0 | 2 | 5 | 8 | 12 |
| \(v_A(t)\) (meters/minute) | 0 | 100 | 40 | -120 | -150 |
Write an integral expression for the position at \(t=12\).
Then use a trapezoidal sum with the three subintervals from the table to approximate that position.
Find the rate (m/min) at which the distance between Train A and Train B is changing at \(t=2\).
Most-appropriate topic codes (CED):
• TOPIC 1.16: Working with the Intermediate Value Theorem (IVT) — part (b)
• TOPIC 6.3: Riemann Sums, Summation Notation, and Definite Integral Notation — part (c)
• TOPIC 3.2: Implicit Differentiation — part (d)
▶️ Answer/Explanation
(a)
Average acceleration on \([2,8]\):
\(\displaystyle \frac{v_A(8)-v_A(2)}{8-2} =\frac{-120-100}{6} =-\frac{220}{6} =\boxed{-\frac{110}{3}\ \text{m/min}^2}.\)
(b)
\(v_A\) is differentiable ⇒ \(v_A\) is continuous on \((5,8)\).
\(v_A(5)=40\) and \(v_A(8)=-120\).
Since \(-100\) lies between \(40\) and \(-120\), the IVT guarantees some \(t\in(5,8)\) with \(v_A(t)=-100\).
\(\boxed{\text{Yes, by IVT on }(5,8).}\)
(c)
Position function: \(\displaystyle s_A(12)=s_A(2)+\int_{2}^{12} v_A(t)\,dt =300+\int_{2}^{12} v_A(t)\,dt.\)
Trapezoidal sum with subintervals \([2,5]\), \([5,8]\), \([8,12]\):
\(\displaystyle \int_{2}^{12}\!v_A(t)\,dt \approx 3\times\frac{100+40}{2}\;+\; 3\times\frac{40+(-120)}{2}\;+\; 4\times\frac{-120+(-150)}{2}\)
\(=3\times 70\;+\;3\times(-40)\;+\;4\times(-135)\)
\(=210-120-540=-450.\)
Therefore \(s_A(12)\approx 300+(-450)=\boxed{-150\ \text{m}}\) (150 m west of the station).
(d)
Let \(x(t)\) be Train A’s east–west position, \(y(t)\) Train B’s north–south position, and \(z(t)\) the distance between them. Then \(z^2=x^2+y^2\).
Differentiate: \(\displaystyle 2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt} \Rightarrow \frac{dz}{dt}=\frac{x\,x’+y\,y’}{z}.\)
At \(t=2\): \(x=300\), \(x’=v_A(2)=100\); \(y=400\), \(y’=v_B(2)=-5(2)^2+60(2)+25=125\);
\(z=\sqrt{300^2+400^2}=500.\)
\(\displaystyle \frac{dz}{dt}=\frac{300\times 100+400\times 125}{500} =\frac{80{,}000}{500} =\boxed{160\ \text{m/min}}.\)
No-Calc Question 5
Most-appropriate topic codes (CED):
• TOPIC 8.10: Disc/Washer About Other Axes — part (b)
• TOPIC 8.13: Arc Length of Smooth Curves — part (c)
▶️ Answer/Explanation
Let \(y_1=xe^{x^{2}}\), \(y_2=-2x\). On \(0\le x\le 1\), \(y_1\ge y_2\).
\(\displaystyle \text{Area}=\int_{0}^{1}\big(y_1-y_2\big)\,dx =\int_{0}^{1}\!\big(xe^{x^{2}}+2x\big)\,dx\)
\(=\left[\tfrac12\,e^{x^{2}}+x^{2}\right]_{0}^{1} =\left(\tfrac12 e+1\right)-\left(\tfrac12+0\right) =\boxed{\tfrac12 e+\tfrac12}.\)
(b)
Rotate about \(y=-2\) (washers; vertical slices).
\(R_{\text{out}}(x)=(y_1+2)\), \(R_{\text{in}}(x)=(2-2x)\).
\(\displaystyle \boxed{V=\pi\int_{0}^{1}\Big[(y_1+2)^{2}-(2-2x)^{2}\Big]\,dx}\), where \(y_1=xe^{x^{2}}\).
(c)
Perimeter \(=\) arc of \(y_1\) on \([0,1]\) \(+\) arc of \(y_2\) on \([0,1]\) \(+\) the vertical segment at \(x=1\).
\(y_1′(x)=e^{x^{2}}(1+2x^{2})\), \(y_2′(x)=-2\).
\(\displaystyle \boxed{ \text{Perimeter}= \int_{0}^{1}\!\sqrt{1+\big(e^{x^{2}}(1+2x^{2})\big)^{2}}\,dx +\int_{0}^{1}\!\sqrt{1+(-2)^{2}}\,dx +(e-(-2)) }\)
\(=\;\int_{0}^{1}\!\sqrt{1+\big(e^{x^{2}}(1+2x^{2})\big)^{2}}\,dx +\sqrt{5} × 1 +(e+2).\)
No-Calc Question 6
Most-appropriate topic codes (CED):
• TOPIC 10.14: Finding Taylor or Maclaurin Series for a Function — part (b)
• TOPIC 10.2: Working with Geometric Series — part (c)
▶️ Answer/Explanation
(a) Radius of convergence \(R\)
Let \(a_n=(-1)^{\,n+1}\dfrac{2^{\,n}}{n}(x-1)^{n}\). Use the Ratio Test:
\[ \left|\frac{a_{n+1}}{a_n}\right| =\left|\frac{(-1)^{n+2}\,\frac{2^{\,n+1}}{n+1}\,(x-1)^{n+1}} {(-1)^{n+1}\,\frac{2^{\,n}}{n}\,(x-1)^{n}}\right| =\frac{n}{n+1}\,\frac{2^{\,n+1}}{2^{\,n}}\,(|x-1|^{\,n+1}\!/|x-1|^{\,n}) =\frac{n}{n+1}\times 2|x-1|. \]
As \(n\to\infty\), \(\dfrac{n}{n+1}\to 1\). Therefore the limit is \(2|x-1|\).
Converges when \(2|x-1|<1 \;\Rightarrow\; |x-1|<\tfrac12\). Hence
\(\boxed{R=\tfrac12}\).
(Endpoint checks determine inclusion in the interval, but do not change \(R\).)
(b) Series for \(f’\) about \(x=1\)
Differentiate term-by-term (valid for \(|x-1|<\tfrac12\)):
\[ \frac{d}{dx}\!\left[(-1)^{\,n+1}\frac{2^{\,n}}{n}(x-1)^{n}\right] =(-1)^{\,n+1}\frac{2^{\,n}}{n}\times n\times (x-1)^{\,n-1} =(-1)^{\,n+1}2^{\,n}(x-1)^{\,n-1}. \]
First three nonzero terms (plug \(n=1,2,3\)):
\(n=1:\; (-1)^{2}2^{1}(x-1)^{0}=2\).
\(n=2:\; (-1)^{3}2^{2}(x-1)^{1}=-4(x-1)\).
\(n=3:\; (-1)^{4}2^{3}(x-1)^{2}=+8(x-1)^{2}\).
Therefore
\(\displaystyle f'(x)=2\;-\;4(x-1)\;+\;8(x-1)^{2}\;-\;\cdots\) with general term \(\displaystyle (-1)^{\,n+1}2^{\,n}(x-1)^{\,n-1}\) for \(n\ge 1\).
(c) Closed forms for \(f’\) and \(f\)
Reindex with \(m=n-1\) so the power starts at \(m=0\):
\[ f'(x)=\sum_{n=1}^{\infty}(-1)^{\,n+1}2^{\,n}(x-1)^{\,n-1} =\sum_{m=0}^{\infty}(-1)^{\,m+2}2^{\,m+1}(x-1)^{m} =2\sum_{m=0}^{\infty}\big(-2(x-1)\big)^{m}. \]
This is geometric with first term \(a_1=2\) and common ratio \(r=-2(x-1)\).
Converges when \(|r|=2|x-1|<1\) (i.e., \(|x-1|<\tfrac12\)). Sum gives
\[ f'(x)=\frac{a_1}{1-r} =\frac{2}{1-(-2(x-1))} =\frac{2}{1+2(x-1)} =\boxed{\frac{2}{2x-1}}. \]
Now integrate to recover \(f\):
\(\displaystyle f(x)=\int \frac{2}{2x-1}\,dx.\)
Let \(u=2x-1\Rightarrow du=2\,dx \Rightarrow dx=\dfrac{du}{2}\). Then
\[ f(x)=\int \frac{2}{u}\cdot\frac{du}{2} =\int \frac{1}{u}\,du =\ln|u|+C =\ln|2x-1|+C. \]
From the original series, \(f(1)=\sum_{n=1}^{\infty}(-1)^{\,n+1}\dfrac{2^{\,n}}{n}(1-1)^{n}=0\).
Hence \(0=f(1)=\ln|2(1)-1|+C=\ln 1 + C \Rightarrow C=0\).
Also, \(|x-1|<\tfrac12 \Rightarrow x\in(\tfrac12,\tfrac32)\Rightarrow 2x-1>0\), so \(|2x-1|=2x-1\).
Therefore \(\boxed{f(x)=\ln(2x-1)}\) for \(|x-1|<\tfrac12\).
(Check: \(\dfrac{d}{dx}\ln(2x-1)=\dfrac{2}{2x-1}=f'(x)\).)